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vodka [1.7K]
2 years ago
8

At a certain point along the path in projectile motion, the projectile has a velocity v whose scalar components are vx = +30 m/s

and vy = +40 m/s. As the projectile moves along the path, what would be its minimum speed?
Physics
2 answers:
Licemer1 [7]2 years ago
6 0
The speed of the projectile would be the magnitude of its horizontal and vertical components combined. Therefore, the minimum speed during its flight will be when it has reached its highest point and the vertical component is 0
At that instant, the speed is:
s = √(30² + 0²)
s = 30 m/s
Nezavi [6.7K]2 years ago
4 0
At the maximum point on the projectile motion path, the object start moving downward due to gravity. Thus at the maximum point, the vertical component of its velocity is zero. Hence only horizontal component of the velocity of project exist. 
So the minimum speed of the project will be 30m/s.
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Two 5.0 mm × 5.0 mm electrodes are held 0.10 mm apart and are attached to 7.5 V battery. Without disconnecting the battery, a 0.
Musya8 [376]

Answer:

A) V = 7.5 V

B) E = 75,000 V/m

C) Q = 16.6 pC

D) V = 7.5 V

E) E = 24,000 V/m

F) Q = 52 pC

Explanation:

Given:

- The Area of plate A = ( 5 x 5 ) mm^2

- The distance between plates d = 0.10 mm

- The thickness of Mylar added t = 0.10 mm

- Voltage supplied by battery V = 7.5 V

Solution:

A) What is the capacitor's potential difference before the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

B) What is the capacitor's electric field before the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 1 (air)                  E = 7.5 / 0.10*10^-3

                                E = 75,000 V/m

C) What is the capacitor's charge Q before the Mylar is inserted?

                                C = k*A*ε / d

k = 1 (air)                   C = ( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 2.213 pF

                                Q = C*V

                                Q = 7.5*(2.213)

                                Q = 16.6 pC

D) What is the capacitor's potential difference after the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

E) What is the capacitor's electric field after the Mylar is inserted?    

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 3.13                     E = 7.5 / (3.13)0.10*10^-3

                                E = 24,000 V/m              

F) What is the capacitor's charge after the Mylar is inserted?      

                                C = k*A*ε / d

k = 3.13                    C = 3.13*( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 6.927 pF

                                Q = C*V

                                Q = 7.5*(6.927)

                                Q = 52 pC                                      

6 0
2 years ago
A roller coaster travels 200 feet horizontally and then rises 135 feet at an angle of 30 degrees above the ground. What is the m
Delvig [45]
Say the initial point is (0,0)

The final point is

x = 200 + 135*cos(30) = 200 + 135*sqrt(3)/2 = 316.91 ft
y = 135*sin(30) = 135/2 = 67.5 ft

Resultant vector = (316.91, 67.5) - (0,0) = 316.91, 67.5) ft
8 0
2 years ago
(d) A beam of white light shines onto a sheet of white paper. An identical beam of light shines onto a mirror. The light is scat
irakobra [83]
QUESTION:-A beam of white light shines onto a sheet of white paper. An identical beam of light shines onto a mirror. The light is scattered from the paper and reflected from the mirror.
Describe how scattering by paper and reflection by a mirror are different from each other.


ANSWER: Scattering sends or reflects light from each point on the object in all directions, whereas reflection sends light from each point on the object in one direction only (or to one point)
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2 years ago
A block of mass m is pushed up against a spring with spring constant k until the spring has been compressed a distance x from eq
Snowcat [4.5K]

Answer:d

Explanation:

Spring is compressed to a distance of x from its equilibrium position

Work done by block on the spring is equal to change in elastic potential energy

i.e. Work done by block W=\frac{1}{2}kx^2

therefore spring will also done an equal opposite amount of work on the block in the absence of external force

Thus work done by spring on the block W=-\frac{1}{2}kx^2

Thus option d is correct

6 0
2 years ago
A circular loop of wire with a radius of 12.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne
Ulleksa [173]

(a) 34 V

The average emf induced in the loop is given by Faraday-Newmann-Lenz law:

\epsilon = -\frac{\Delta \Phi_B}{\Delta t} (1)

where

\Delta \Phi_B is the variation of magnetic flux through the coil

\Delta t = 2.0 ms = 0.002 s is the time interval

We need to find the magnetic flux before and after. The magnetic flux is given by:

\Phi_B = BA

where

B is the magnetic field intensity

A is the area of the coil

The radius of the coil is r = 12.0 cm = 0.12 m, so its area is

A=\pi r^2 = \pi (0.12 m)^2 = 0.045 m^2

At the beginning, the magnetic field is

B_i = 1.5 T

so the flux is

\Phi_i = B_i A = (1.5 T)(0.045 m^2)=0.068 Wb

while after the removal of the coil, the magnetic field is zero, so the flux is also zero:

\Phi_f = 0

so the variation of magnetic flux is

\Delta \Phi = 0-0.068 Wb=-0.068 Wb

And substituting into (1) we find the average emf in the coil

\epsilon=-\frac{-0.068 Wb}{0.002 s}=34 V

(b) Counterclockwise

In order to understand the direction of the induced current, we have to keep in mind the negative sign in Lenz's law (1), which tells that the direction of the induced current must be such that the magnetic field produced by this current opposes the variation of magnetic flux in the coil.

In this situation, the magnetic flux through the coil is decreasing, since the coil is removed from the field. So, the induced current must be such that it produces a magnetic field whose direction is the same as the direction of the external magnetic field, which is upward along the positive z-direction.

Looking down from above and using the right-hand rule on the loop (thumb: direction of the current, other fingers wrapped: direction of magnetic field), we see that in order to produce at the center of the coil a magnetic field which is along positive z-direction, the induced current must be counterclockwise.

4 0
2 years ago
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